Physics2 Lesson 1 Chapter Review I

Fluids and Pressure

Chapter Review 

(source: http://cwx.prenhall.com/bookbind/pubbooks/walker2/chapter15/custom1/deluxe-content.html)

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In this chapter we study fluids.

--> Fluids are characterized by their ability to flow; both liquids and gases are considered to be fluids. An understanding of fluid behavior is essential to life and applications of this understanding are essential to many of the conveniences of modern living.

15-1 - 15-2   Density and Pressure

One of the more convenient properties used to describe a fluid is its density . The density of a substance is a measure of how compact the substance is; that is, how much mass is packed into a volume of the substance. On average the density of a substance is the amount of mass M divided by the volume V taken up by that mass

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Another important quantity in the study of fluids is pressure P. On the average, the pressure that is applied to an object is the amount of force F (normal to the surface) divided by the area A over which the force spreads

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As the above expressions indicates, the units of pressure are those of force divided by area; in SI units this is called the pascal (Pa): 1 Pa = 1 N/m². An important property of the pressure in a fluid is that it is equally applied in all directions (at a given depth) and applies forces that are perpendicular to any surface in the fluid.

When considering the pressure on or within a fluid it is important to recognize that for many applications we must account for a constant atmospheric pressure (P_at = 1.01 x 10⁵ Pa). Because of the constant presence of the atmosphere we are often only interested in the pressure above and beyond that applied by the atmosphere. This additional pressure is called the gauge pressureP_g

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where P is the total pressure applied (sometimes called the absolute pressure).

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Exercise 15.1 The Density of a Fluid: If 1.00 gallons of a certain fluid weighs 3.22 lb, what is its density?

Solution: We are given the following information:

Given: V = 1.00 gal, W = 3.22 lb   Find: 

Since the density is given by  = m/V we need to find the mass from the weight and convert everything to SI units. To get the mass we have

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The volume can be immediately converted to give

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We are now ready to calculate the density as

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Example 15.2 Gauge Pressure of Water: A uniform cylindrical container has a radius of 7.8 cm and a height of 13.2 cm. If this container is completely filled with water, what gauge pressure does the water apply to the bottom of the container?

Picture the Problem The picture shows a cylindrical container filled with water.

Strategy The pressure applied by the water will be the force that the water applies (equal to its weight) divided by the bottom area of the cylinder.

Solution

1. The volume of water equals the volume of the cylinder:
2. Determine mass of the water
3. Determine the weight of the water:
4. Obtain the gauge pressure:

Insight The above answer is the gauge pressure because we have ignored atmospheric pressure.

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Practice Quiz

If it takes as much volume of fluid 1 to weigh the same as fluid 2, how do their densities compare?  fluid 1 is twice as dense as fluid 2  fluid 1 is half as dense as fluid 2  fluid 1 and fluid 2 have equal densities  fluid 1 is four times less dense as fluid 2  [None of the above]

If a cylinder of height 0.850 m and radius 0.250 m is filled with water, what is the total pressure on the bottom of the cylinder?  8.34 kPa  133 kPa  2.08 kPa  109 kPa  46.2 kPa

15-3   Static Equilibrium in Fluids: Pressure and Depth

In this and the next several sections we will consider the properties of static fluids, that is, fluids that do not flow. In the case of static fluids, every part of the fluid and every object within the fluid is in static equilibrium. One of the basic properties of fluids that is very important to our ability to understand fluid behavior is known as Pascal's principle:

External pressure applied to an enclosed fluid is transmitted unchanged throughout the fluid.

Pascal's principle is important in determining the dependence of pressure on the depth within a fluid. Without any external pressure applied to the outer surface of a fluid, the pressure measured at a depth h beneath the surface arises from the weight of the fluid above the given level. The amount of this increase in pressure is given by gh. If there is external pressue on the fluid, such as from the atmosphere and/or any other source, this pressure is transmitted undiminished to every point in the fluid and it must be added to the pressure due to the weight of the fluid. Thus, for the dependence of pressure on depth we have

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where P₂ is the pressure at a given level within the fluid and P₁ is the pressure at a height h above that level.

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Example 15.3 Blood Pressure with Depth: Human blood has a density of approximately 1.05 x 10³ kg/m³. Use this information to estimate the difference in blood pressure between the brain and the feet in a person who is approximately six feet tall.

Picture the Problem The picture shows a person approximately six feet in height.

Strategy We attempt this approximation to two significant figures by applying the above result for the dependence of pressure on depth using blood as the fluid.

Solution

1. Convert the height to meters:
2. The difference in pressure is given by:
3. Obtain the numerical result:

Insight This is only an estimate for many reasons. The blood in the body is not a static fluid, but it flows; and between the head and the feet there is a pump (the heart) which will affect the result. If you plan to study medicine, see if you can find out the difference in blood pressure between the head and the feet.

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Pascal's principle is also key to understanding the hydraulic lift. This device uses fluid pressure to convert a small input force into a large output force. The input force F₁ is applied to a fluid over a comparatively small area A₁ giving rise to a pressure change of F₁/A₁ = F₂/A₂, we must get a larger output force F₂ > F₁ if it is spread over a larger area A₂ > A₁. Therefore,

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The consequence of getting this larger output force is that the distance through which this force can move an object at the output, d₂, is smaller than the distance at the input d₁. Since the same volume of fluid moves at the input and output, A₁d₁ = A₂d₂, the output distance is given by

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Exercise 15.4 Hydraulic Lift: Your job at Dave's Manufacturing Company is to design a hydraulic lift for a client. This client typically needs to raise objects through a height of 0.500 meters. The system should be easily used by people of average height so the input distance should not exceed 5.00 ft. What would be a good ratio of output area to input area to consider?

Solution: We are given the following information.

Given: d₁ = 5.00 ft, d₂ = 0.500 m   Find: A₂/A₁

From the information given, we know that the ratio of the input distance to the output distance is directly proportional to the ratio we seek

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Therefore, a ratio of

The output area should be at least 3.05 times greater than the input area. Of course, in a more realistic situation you'd want more information from the client. What additional information might you want?

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Practice Quiz

What is the gauge pressure 3.4 m below the surface of a container filled with a fluid of density 550 kg/m3?  33 kPa  1900 Pa  18 kPa  100 kPa  550 Pa

The gauge pressure at a particular location in a fluid of density 870 kg/m3 is 120 kPa. What is the gauge pressure in the fluid 5.9 m above this location?  70 kPa  50 kPa  62 kPa  58 kPa  20 kPa

Physics2 Lesson 1- Fluids

Fluids

(source: Cliffs Notes)

A fluid is a substance that cannot maintain its own shape but takes the shape of its container. Fluid laws assume idealized fluids that cannot be compressed.

Density and pressure

The density (ρ) of a substance of uniform composition is its mass per unit volume: ρ = m/ V. In the SI system, density is measured in units of kilograms per cubic meter.

Imagine an upright cylindrical beaker filled with a fluid. The fluid exerts a force on the bottom of the container due to its weight. Pressure is defined as the force per unit area: P = F/ A , or in terms of magnitude, P = mg/A, where mg is the weight of the fluid. The SI unit of pressure is N/m², called a pascal. The pressure at the bottom of a fluid can be expressed in terms of the density (ρ) and height (h) of the fluid: 

or P = ρ hg. The pressure at any point in a fluid acts equally in all directions. This concept is sometimes called the basic law of fluid pressure.

Pascal's principle

Pascal's principle may be stated thus: The pressure applied at one point in an enclosed fluid under equilibrium conditions is transmitted equally to all parts of the fluid. This rule is utilized in hydraulic systems. In Figure 1 , a push on a cylindrical piston at point a lifts an object at point b.

Figure 1 Pascal's principle is used to easily lift a car.

Let the subscripts a and b denote the quantities at each piston. The pressures are equal; therefore, P _a = P _b . Substitute the expression for pressure in terms of force and area to obtain f _a / A _a = ( F _b / A _b ). Substitute π r² for the area of a circle, simplify, and solve for F _b : F _b =( F _a )( r _b ²/ r _a ²). Because the force exerted at point a is multiplied by the square of the ratio of the radii and r _b > r _a , a modest force on the small piston a can lift a relatively larger weight on piston b.

Archimedes' principle

Water commonly provides partial support for any object placed in it. The upward force on an object placed in a fluid is called the buoyant force. According toArchimedes' principle, the magnitude of a buoyant force on a completely or partially submerged object always equals the weight of the fluid displaced by the object.

Archimedes' principle can be verified by a nonmathematical argument. Consider the cubic volume of water in the container of water shown in Figure 2 . This volume is in equilibrium with the forces acting on it, which are the weight and the buoyant force; therefore, the downward force of the weight ( W) must be balanced by the upward buoyant force ( B), which is provided by the rest of the water in the container.

Figure 2 Weight is balanced by buoyant force within a volume of water.

If a solid floats partially submerged in a liquid, the volume of liquid displaced is less than the volume of the solid. Comparing the density of the solid and the liquid in which it floats leads to an interesting result. The formulas for density are D s = m s/ V s and D l = m l / V l , where D is the density, V is the volume, m is the mass, and the subscripts s and l refer to quantities associated with the solid and the liquid respectively. Solving for the masses leads to m s = D s V s and m l = D l V l . According to Archimedes' principle, the weights of the solid and the displaced liquid are equal. Because the weights are simply mass times a constant (g), the masses must be equal also; therefore, D s V s = D l V l or D s / D l = V l V l . Now, V = Ah, where A is the cross-sectional area and h is the height. For a solid floating in liquid,A l = A s and h l is the height of the solid that is submerged, hsub. With these substitutions, the above relationship becomes D s / D l = hsub/ hs; therefore, the fractional part of the solid that is submerged is equal to the ratio of the density of the solid to the density of the surrounding liquid in which it floats. For example, about 90 percent of an iceberg is beneath the surface of sea water because the density of ice is about nine-tenths that of sea water.

Bernoulli's equation

Imagine a fluid flowing through a section of pipe with one end having a smaller cross-sectional area than the pipe at the other end. The flow of liquids is very complex; therefore, this discussion will assume conditions of the smooth flow of an incompressible fluid through walls with no drag. The velocity of the fluid in the constricted end must be greater than the velocity at the larger end if steady flow is maintained; that is, the volume passing per time is the same at all points. Swiftly moving fluids exert less pressure than slowly moving fluids. Bernoulli's equation applies conservation of energy to formalize this observation: P + (1/2) ρ v² + ρ gh= a constant. The equation states that the sum of the pressure (P), the kinetic energy per unit volume, and the potential energy per unit volume have the same value throughout the pipe.

Physics 2 Lecture 1 - Fluids and Pressure

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Topic covered is about Fluids and Pressure.

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